Electrostatic Potential and Capacitance - Result Question 45

47. If the potential of a capacitor having capacity 6μF is increased from 10V to 20V, then increase in its energy will be

(a) 4×104J

(b) 4×104J

(c) 9×104J

(d) 12×106J

[1995]

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Answer:

Correct Answer: 47. (c)

Solution:

  1. (c) Capacitance of capacitor (C)=6μF=6 ×106F; Initial potential (V1)=10V and final potential (V2)=20V.

The increase in energy (ΔU)

=12C(V22V12)

=12×(6×106)×[(20)2(10)2]

=(3×106)×300=9×104J.



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